The output voltage of an RC integrator circuit is proportional to the integral of the input voltage.

This means that a triangular wave should transform into a quadratic curve (since the integral of a linear function is a parabola).

Why does it convert a triangular wave into a sine wave then? What am I missing here?

enter image description here

  • 1
    \$\begingroup\$ Please include a schematic. Part values and the triangle wave frequency would help, too, because an RC circuit isn't a pure integrator -- it's a low-pass filter. \$\endgroup\$ – TimWescott Dec 29 '20 at 20:40
  • 3
    \$\begingroup\$ It converts a triangle wave into a parabolic wave, which is "close enough" to a sine wave. \$\endgroup\$ – Hearth Dec 29 '20 at 20:42
  • 4
    \$\begingroup\$ get glasses, and look closer. Output may appear sinusoidal - but it isn't. Doing a Fourier transform on the output will show some harmonic content. \$\endgroup\$ – glen_geek Dec 29 '20 at 20:42
  • \$\begingroup\$ You can also view this as a low-pass filter. The more integrators you put in series, the better the high frequencies are filtered out, and the more the lowest one dominates. \$\endgroup\$ – user253751 Dec 29 '20 at 20:48
  • 2
    \$\begingroup\$ Any non-sinusoidal signal is composed out of multiple sinusoidal signals in superposition (harmonics). The more you filter those harmonics away with low pass filters the closer you get to the fundamental sinewave. \$\endgroup\$ – Unimportant Dec 29 '20 at 20:59

So, a square wave has harmonics at all odd multiples of the fundamental, with strength diminishing as the number of the harmonic:

$$x_s(t) \simeq \sum_{n = 0}^\infty \frac{\cos 2 \pi (2n + 1) t}{2n + 1}$$

A triangle wave is just the square wave, integrated, with the appropriate constant added to make things tidy:

$$x_\Delta(t) \simeq \sum_{n = 0}^\infty \frac{\sin 2 \pi (2n + 1) t}{(2n + 1)^2}$$

A parabolic wave is the same thing, again:

$$x_p(t) \simeq \sum_{n = 0}^\infty -\frac{\cos 2 \pi (2n + 1) t}{(2n + 1)^3}$$

Because of that cubic in the denominator, the difference between a sine wave and this "pseudo-sine" wave is very small. And it turns out that it's pretty hard to see the deviation on a graph anyway.

If you plot the quadratic pseudo-sine wave on a graph, superimposed on a real sine wave, you'll see the (slight) difference.


It doesn't. It converts a triangle wave into a parabolic wave, which is "close enough" to a sine wave. The image you provided is making an approximation.

That said, an RC circuit isn't an integrator, but a low-pass filter. The two are similar, but not identical.

  • \$\begingroup\$ Well to be fair, after the pole, and RC LP behaves just like an integrator. If the triangular wave in the picture is out of band enough, you will get a sinusoidal wave. \$\endgroup\$ – Vladimir Cravero Dec 30 '20 at 11:25
  • \$\begingroup\$ I agree with @Hearth: These circuits ARE NOT integrators, whichever close to an integration operation could be the relationship between input and output. Naming things with wrong names can lead only to confusion. \$\endgroup\$ – Orestes Mas Dec 30 '20 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.